In the given figure,the potentiometer wire $AB$ has a resistance of $5 \, \Omega$ and a length of $10 \, m$. The balancing length $AM$ for the $emf$ of $0.4 \, V$ is ............... $m$.

The length of a potentiometer wire is $\ell$. $A$ cell of emf $E$ is balanced at a length $\ell/3$ from the positive end of the wire. If the length of the wire is increased by $\ell/2$,at what distance will the same cell give a balanced point?

In the determination of the internal resistance of a cell with a potentiometer,the error in the measurement of the balancing length is $\pm 1 \text{ mm}$. When the cell alone is connected in the circuit,the balancing length is obtained at $60 \text{ cm}$ and when the cell is shunted with a resistance of $10 \Omega \pm 2 \%$,the balancing length is obtained at $50 \text{ cm}$. The error in the determination of the internal resistance of the cell is (in $\%$)

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

$A$ null point is obtained at $200 \ cm$ on a potentiometer wire when a cell in the secondary circuit is shunted by $5 \ \Omega$. When a resistance of $15 \ \Omega$ is used for shunting,the null point moves to $300 \ cm$. The internal resistance of the cell is: (in $Omega$)

While doing an experiment with a potentiometer as shown in the figure,it was found that the deflection is one-sided and $(i)$ the deflection decreased while moving the jockey from one end $A$ of the wire to the end $B$; $(ii)$ the deflection increased while the jockey was moved towards the end $B$.
$(i)$ Which terminal ($+$ or $-ve$) of the cell $E_1$ is connected at $X$ in case $(i)$ and how is $E_1$ related to $E$?
$(ii)$ Which terminal of the cell $E_1$ is connected at $X$ in case $(ii)$?